Students can use CBSE Previous Year Question Papers Class 12 Chemistry with Solutions and CBSE Class 12 Chemistry Question Paper 2023 (Series HFGIE/2) to familiarize themselves with the exam format and marking scheme.
CBSE Class 12 Chemistry Question Paper 2023 (Series: HFGIE/2) with Solutions
Time Allowed: 3 hours
Maximum Marks: 70
General Instructions:
- This Question Paper contains 35 questions. All questions are compulsory.
- Question paper is divided into Five sections-A, B, C, D and E.
- In Section-A: Question Numbers 1 to 18 are Multiple Choice (MCQs) type questions carrying 1 mark each.
- In Section-B: Question Numbers 19 to 25 are Very Short Answer (VSA) type questions carrying 2 marks each.
- In Section-C: Question Numbers 26 to 30 are Short Answer (SA) type questions carrying 3 marks each.
- In Section-D: Question Numbers 31 & 32 are case based questions carrying 4 marks each.
- In Section-E: Question Numbers 33 to 35 are Long Answer (LA) type questions carrying 5 marks each.
- There is no overall choice. However, an internal choice has been provided in 2 questions in Section-B, 2 questions in Section-C, 2 questions in Section-D and 2 questions in Section-E.
- Use of calculators is NOT allowed.
Set I Code No. 56/2/1
Section-A
Question 1.
The conversation of an alkyl halide into an alkene by alcoholic KOH is classified as [1]
(a) a substitution reaction
(b) an addition reaction
(c) a dehydrohalogenation reaction
(d) a dehydration reaction
Answer:
(c) a dehydrohalogenation reaction
When ethyl chloride is heated in presence of alcoholic KOH, it undergoes dehydro- halogenation to give alkene product.
Question 2.
The oxidation state of Fe in [Fe(CO)
5
] is [1]
(a) +2
(b) 0
(c) +3
(d) +5
Answer:
(b) 0
Carbonyl complexes have zero oxidation state.
Question 3.
Among the following, which is the strongest base? [1]
Answer:
(c)
In benzylamine, the amino group is not directly attached to the benzene ring. Hence, electrons cannot participate in resonance and the electrons on nitrogen are available for donation.
Question 4.
The slope in the plot of ln[R] vs. time for a first order reaction is
(a) \(\frac{+k}{2.303}\)
(b) -k
(c) \(\frac{-k}{2.303}\)
(d) +k
Answer:
(c) \(\frac{-k}{2.303}\)
Question 5.
An α-helix is a structural feature of [1]
(a) Sucrose
(b) Folypeptides
(c) Nucleotides
(d) Starch
Solution:
(b) Folypeptides
The α-helix is a structural feature of proteins which are polypeptides chains of amino acids.
Question 6.
Racemisation occurs in [1]
(a) S
N
1 reaction
(b) S
N
2 reaction
(c) Neither S
N
1 nor S
N
2 reaction
(d) S
N
2 reaction as well as S
N
1 reaction
Answer:
(a) S
N
1 reaction
Racemization occurs in S N 1 reaction only as the sp 2 hybridized carbocation is formed as intermediate which could be attacked by nucleophile from both front and back side equally to form racemic mixture.
Question 7.
Value of Henry’s constant K
H
: [1]
(a) increases with decrease in temperature
(b) decreases with increase in temperature
(c) increases with increase in temperature
(d) remains constant
Answer:
(c) increases with increase in temperature
Question 8.
Which of the following solutions of KCl will have the highest value of molar conductivity? [1]
(a) 0.01 M
(b) 1 M
(c) 0.5 M
(d) 0.1 M
Answer:
(a) 0.01 M
∵ Λ
m
= \(\frac{1000 \times k}{\mathrm{M}}\) ⇒ Λ
m
∝ \(\frac{1}{\mathrm{M}}\)
Molar conductivity increases with decrease in molarity.
Question 9.
Which of the following reactions are feasible? [1]
(a) CH
3
CH
2
Br + Na
+
O
–
C(CH
3
)
3
→ CH
3
CH
2
—O—C (CH
3
)
3
(b) (CH
3
)
3
C—Cl + Na
+
O
–
CH
2
CH
3
→ CH
3
CH
2
—O—C (CH
3
)
3
(c) Both (a) and (b)
(d) Neither (a) nor (b)
Answer:
(a) CH
3
CH
2
Br + Na
+
O
–
C(CH
3
)
3
→ CH
3
CH
2
—O—C (CH
3
)
3
For preparing unsymmetrical ethers the halide should be primary because Williamson synthesis occurs by S N 2 mechanism and primary alkyl halides are most reactive in S N 2 reactions.
Question 10.
Which of the following is most reactive in nucleophilic addition reactions? [1]
(a) HCHO
(b) CH
3
CHO
(c) CH
3
COCH
3
(d) CH
3
COC
2
H
5
Answer:
(a) HCHO
Presence of alkyl group decreases the reactivity by decreasing the electron deficiency. Hence, HCHO is more reactive towards nucleophilic addition reactions.
Question 11.
Which of the following does not give aldol condensation reaction? [1]
(a) CH
3
—CHO
(b)
(c)
(d) CH
3
COCH
3
Answer:
(c)
Benzaldehyde will not give aldol condensation due to absence of α—H atom.
Question 12.
For the reaction 3A → 2B, rate of reaction +\(\frac{d[\mathrm{~B}]}{d t}\) is equal to
(a) \(\frac{-3}{2} \frac{d[\mathrm{~A}]}{d t}\)
(b) \(\frac{-2}{3} \frac{d[\mathrm{~A}]}{d t}\)
(c) \(\frac{-1}{3} \frac{d[\mathrm{~A}]}{d t}\)
(d) \(+\frac{2 d[\mathrm{~A}]}{d t}\)
Answer:
(b) \(\frac{-2}{3} \frac{d[\mathrm{~A}]}{d t}\)
For 3A → 2B
Rate of reaction = \(\frac{-1}{3} \frac{d[\mathrm{~A}]}{d t}\) = \(+\frac{1}{2} \frac{d[\mathrm{~B}]}{d t}\) ∴ \(\frac{d[\mathrm{~B}]}{d t}\) = \(-\frac{2}{3} \frac{d[\mathrm{~A}]}{d t}\)
Question 13.
Which of the following characteristics of transition metals is associated with their catalytic activity? [1]
(a) Paramagnetic nature
(b) Colour of hydrated ions
(c) High enthalpy of atomisation
(d) Variable oxidation states
Answer:
(d) Variable oxidation states
Due to formation of variable oxidation states, transition metals show catalytic property.
Question 14.
The formula of the complex dichloridobis (ethane -1, 2-diamine) platinum (IV) nitrate is [1]
(a) [Pt Cl
2
(en)
2
(NO
3
)
2
]
(b) [Pt Cl
2
(en)
2
] (NO
3
)
2
(c) [Pt Cl
2
(en)
2
(NO
3
)] NO
3
(d) [Pt (en)
2
(NO
3
)
2
] Cl
2
Answer:
(b) [Pt Cl
2
(en)
2
] (NO
3
)
2
Given below are two statements labelled as Assertion (A) and Reason (R). Select the most appropriate answer from the options given below:
(a) Both (A) and (R) are true and (R) is the correct explanation of (A).
(b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
(c) (A) is true, but (R) is false.
(d) (A) is false, but (R) is true.
Question 15.
Assertion (A): Osmotic pressure is a colligative property.
Reason (R) : Osmotic pressure is proportional to the molality. [1]
Answer:
(b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
Osmotic pressure is a colligative property as it depends on the number of solute present and not on the nature of solute.
Osmotic pressure is directly proportional to molality and temperature, i.e., π = CRT.
Question 16.
Assertion (A): Conductivity decreases with the decrease in concentration of electrolyte.
Reason (R) : Number of ions per unit volume that carry the current in a solution decreases on dilution.
Answer:
(a) Both (A) and (R) are true and (R) is the correct explanation of (A).
Question 17.
Assertion (A): Copper is a non-transition element.
Reason (R) : Copper has completely filled d-orbitais in its ground state. [1]
Answer:
(d) (A) is false, but (R) is true.
Copper is a transition element due to partially filled d-orbitals (d 9 – Cu 2+ ) in its ionic state.
Question 18.
Assertion (A): Nucleophilic substitution of iodoetharie is easier than chloroethane.
Reason (R) : Bond enthalpy of C-I bond is less than that of C-Cl bond. [1]
Answer:
(a) Both (A) and (R) are true and (R) is the correct explanation of (A).
Section – B
Question 19.
The vapour pressure of pure liquid X and pure liquid Y at 25 °c are 120 mm Hg and 160 mm Hg respectively. If equal moles of X and Y are mixed to form an ideal solution, calculate the vapour pressure of the solution. [2]
Answer:
Suppose the moles of X and Y are x. Hence, their mole fraction = \(\frac{x}{2 x}\) = \(\frac{1}{2}\).
Given, vapour pressure of pure liquid ‘X’ = 120 mm Hg
Vapour pressure of pure liquid ‘Y’ = 160’mm Hg
Question 20.
(a) Give reasons: [2 × 1]
(i) Mercury cell delivers a constant potential during its life time.
Answer:
Mercury cell delivers a constant potential during its lifetime because the overall reaction does not involve any ion in solution whose concentration can change.
(ii) In the experimental determination of electrolytic conductance, Direct Current (DC) is not used. [2]
Answer:
Because DC (Direct Current) results in the electrolysis of the electrolytic solution. As a result, concentration of the electrolyte near the electrodes changes and these result in change in the resistance of the solution.
Or
(b) Define fuel cell with an example. What advantages do the fuel cells have over primary and secondary batteries? [2]
Answer:
Fuel cell. These are galvanic cells in which the energy of combustion of fuels like hydrogen, methane, etc. is directly converted into electrical energy, e.g., H
2
—O
2
fuel cell.
Advantages of fuel cells over primary and secondary batteries are:
- High efficiency. They are more efficient than conventional method of generating electricity.
- There is no electrode required in it as in an ordinary battery. The fuel can be fed continuously to produce power.
- They do not cause pollution.
Question 21.
(a) The conversion of molecule A to B followed second order kinetics. If concentration of A increased to three times, how will it affect the rate of formation of B?
Answer:
For second order kinetics, we have
r = K[A]
2
When concentration of A increased to 3 times.
The rate of formation of B will be r = K[3A]
2
= 9 times
So, the rate of formation of B increases by 9 times.
(b) Define Pseudo first order reaction with an example. [2 × 1]
Answer:
Pseudo First Order Reaction. The reactions which are not truly of first order but under certain conditions become reactions of first order are called Pseudo first order reactions. Example. Hydrolysis of ethyl acetate
Due to high amount of water, its concentration remains practically unchanged and rate of reaction depends only on concentration of ethyl acetate and hence order is 1.
Question 22.
(a) Write the IUPAC names of the following: [2 × 1]
(i) [Co(NH
3
)
5
(ONO)]
2+
Answer:
IUPAC name of [Co(NH
3
)
5
(ONO)]
2+
is Pentaaminenitrito-O-cobalt (III) ion.
(ii) K
2
[NiCl
4
]
Answer:
IUPAC name of K
2
[NiCl
4
] is Potassium tetrachloridonickelate (II).
Or
(b)
(i) What is a chelate complex? Give one example. [2 × 1]
Answer:
When a bidentate or a polydentate ligand binds through its 2 or more donor atoms with same central metal atom or ion then a ring structured complex is formed which is known as chelate complex.
Example.
(ii) What are heteroleptic complexes? Give one example.
Answer:
The complexes in which the metal is bound to more than one kind of donor groups or ligands are called heteroleptic complexes.
Example. [NiCl
2
(H
2
O)
4
] Tetraaquodichloridonickel (II)
This complex contains 2 types of ligands viz., Cl and H
2
O.
Question 23.
Write the chemical equation involved in the following reactions: [2 × 1]
(a) Reimer-Tiemann reaction
Answer:
Reimer-Tiemann reaction:
(b) Acetylation of Salicylic acid
Answer:
Acetylation of Salicyclic acid:
Question 24.
Do the following conversions in not more than two steps: [2 × 1]
(a)
Answer:
(b)
Answer:
Question 25.
Write two differences between DNA and RNA. [1 × 2]
Answer:
Difference between DNA and RNA:
| DNA | RNA |
| 1. The sugar present in DNA is 2-deoxy-(-) ribose. | The sugar present in RNA is D-(-)ribose. |
| 2. DNA contains cytosine and thymine as pyrimidine bases. | RNA contains cytosine and uracil as pyrimidine bases. |
Section – C
Question 26.
(a)
(i) Write the mechanism of the following reaction: [2 + 1]
Answer:
Mechanism:
(ii) Why ortho-nitrophenol is steam volatile while para-nitrophenol is not?
Answer:
o-nitrophenol is steam volatile than p-nitrophenol due to presence of intra-molecular hydrogen bonding in it.
Or
(b) What happens when: [3 × 1]
(i) Anisole is treated with CH
3
Cl/ anhydrous AlCl
3
?
(ii) Phenol is oxidised with Na
2
Cr
2
O
7
/H
+
?
(iii) (CH
3
)
3
C —OH is heated with Cu/573 K?
Write chemical equation in support of your answer.
Answer:
Question 27.
Answer any three of the following: [3 × 1]
(a) Which isomer of C
5
H
10
gives a single monochloro compound C
5
H
9
Cl in bright sunlight?
Answer:
C
5
H
10
is an alkene so it must be Pentene. Allylic chlorination occurs and hydrogen adjacent to the C = C will be substituted.
(b) Arrange the following compounds in increasing order of reactivity towards S
N
2 reaction:
2-Bromopentane, 1-Bromopentane, 2-Bromo-2-methylbutane
Answer:
2 -Bromo-2-methylbutane < 2-Bromopentane < 1-Bromopentane
(c) Why p-dichlorobenzene has higher melting point than those of ortho- and meta-isomers?
Answer:
p-isomers are comparatively more symmetrical and fit closely in the crystal lattice, thus require more heat to break these strong forces of attraction. Therefore higher melting point than o- and m-isomers.
(d) Identify A and B in the following:
Answer:
Question 28.
A first order reaction is 50% complete in 30 minutes at 300 K and in 10 minutes at 320 K. Calculate activation energy (E
a
) for the reaction. [R = 8.314 JK
-1
mol
-1
] [3]
[Given: log 2 = 0.3010, log 3 = 0.4771, log 4 = 0.6021]
Answer:
Using t
1/2
= \(\frac{0.693}{K}\)
At 300K, K
1
= \(\frac{0.693}{30}\) = 0.0231 min
-1
; At 320 K, K
2
= \(\frac{0.693}{10}\) = 0.0693 min
-1
Using Arhenius equation, log\(\left(\frac{\mathrm{K}_2}{\mathrm{~K}_1}\right)\) = \(\frac{\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{R}}\left(\frac{\mathrm{T}_2-\mathrm{T}_1}{\mathrm{~T}_1 \mathrm{~T}_2}\right)\)
log\(\left(\frac{0.0693}{0.0231}\right)\) = \(\frac{\mathrm{E}_{\mathrm{a}}}{2.303 \times 8.314 \times 10^{-3} \mathrm{KJ} \mathrm{mol}^{-1} \mathrm{~K}^{-1}}\left(\frac{320-300}{300 \times 320}\right)\) ∴ E
a
= 43.85 KJ mol
-1
Question 29.
When 19.5 g of F —CH
2
—COOH (Molar mass = 78 g mol
-1
), is dissolved in 500 g of water, the depression in freezing point is observed to be 1°C. Calculate the degree of dissociation of F —CH
2
—COOH. [Given: K
f
for water = 1.86 K kg mol
-1
] [3]
Answer:
Given: Molar mass of F—CH
2
—COOH = 78 g/mol
Amount of solvent water = 500 g
K
f
= 1.86 K kg mol
-1
, Number of moles = 19.5 g
Number of moles of F—CH
2
—COOH = \(\frac{19.5}{78}\) = 0.25
Molality, m \(=\frac{\text { No. of moles }}{1 \mathrm{~kg} \text { of solvent }}\) = \(\frac{0.25}{500}\) × 1000 = 0.50 m
Calculated freezing point depression
Question 30.
(a) Draw the geometrical isomers of [Co(en)
2
Cl
2
]
2+
. Which geometrical isomer of [Co(en)
2
Cl
2
]
2+
is not optically active and why?
Answer:
Geometrical isomers of [Co(en)
2
Cl
2
]
2+
are:
Transform is not optically active because it has a plane of symmetry.
(b) Write the hybridisation and magnetic behaviour of [CoF
6
]
3-
.
[Given: Atomic number of Co = 27] [2 + 1]
Answer:
[CoF
6
]
3-
—Hexafluorido cobalt (III) ion.
Here F is a weak field ligand and is unable to do pairing.
Co
3+
= [Ar] 3d
6
4
0
4p
0
4d
0
Due to presence of 4 unpaired electrons, it is paramagnetic in nature.
Section – D
The following questions are Case Based Questions. Read the passage carefully and answer the questions that follow:
Question 31.
The carbon-oxygen double bond is polarised in aldehydes and ketones due to higher electronegativity of oxygen relative to carbon. Therefore they undergo nucleophilic addition reactions with a number of nucleophiles such as HCN, NaHSO
3
, alcohols, ammonia derivatives and Grignard reagents. Aldehydes are easily oxidised by mild oxidising agents as compared to ketones. The carbonyl group of carboxylic acid does not give reactions of aldehydes and ketones. Carboxylic acids are considerably more acidic than alcohols and most of simple phenols.
Answer the following:
(a) Write the name of the product when an aldehyde reacts with excess alcohol in presence of dry HCl. [1]
Answer:
(b) Why carboxylic acid is a stronger acid than phenol? [1]
Answer:
Because the carboxylate ion (RCOOH \(\rightleftharpoons\) RCOO
–
+ H
+
) is relatively more stable as compared to phenoxide ion. Thus, release of H
+
ion from carboxylic acid is comparatively easier and it behaves as a stronger acid than phenol.
(c) (i) Arrange the following compounds in increasing order of their reactivity towards CH
3
MgBr:
Answer:
(ii) Write a chemical test to distinguish between propanal and propanone. [2 × 1]
Answer:
Propanone on reacting with I
2
and NaOH gives a yellow ppt of iodoform but propanal does not respond to this test.
Or
(c) Write the main product in the following:
(i)
(ii)
Answer:
Question 32.
Carbohydrates are optically active polyhydroxy aldehydes and ketones. They are also called saccharides. All those carbohydrates which reduce Fehling’s solution and Tollen’s reagent are referred to as reducing sugars. Glucose, the most important source of energy for mammals, is obtained by the hydrolysis of starch. Vitamins are accessory food factors required in the diet. Proteins are the polymers of a-amino acids and perform various structural and dynamic functions in the organisms. Deficiency of vitamins leads to many diseases.
Answer the following:
(a) The penta-acetate of glucose does not react with Hydroxylamine. What does it indicate?
Answer:
Acetylation of glucose with acetic anhydride gives glucose pentaacetate which confirms the presence of five —OH groups which are attached to different ‘C’ atoms.
(b) Why cannot vitamin C be stored in our body? [1]
Answer:
Vitamin C is a water soluble vitamin which is eliminated from our body in the form of urine.
(c) Define the following as related to proteins: [2 × 1]
(i) Peptide linkage
Answer:
Peptide linkage. It is an amide linkage formed between —COOH group of one α-amino acid and NH
2
group of the other α-amino acid by loss of a molecule of water.
— CO — NH — bond is called Peptide linkage.
(ii) Denaturation
Answer:
Denaturation. Due to coagulation of globular protein under the influence of change in temperature, change in pH etc., the native shape of the protein is destroyed and biological activity is lost and the protein so formed is called denaturated proteins and the phenomenon is denaturation.
Or
(c) Define the following as related to carbohydrates: [2 x 1]
(i) Anomers
Answer:
Anorners. A pair of stereoisomers which differ in configuration only around C
1
are called anomers. Two isomers are said to be anomers if the isomerisation in the molecule is at first carbon.
(ii) Glycosidic linkage
Answer:
Glycosidic linkage. The two monosaccharide units are joined together through an etheral or oxide linkage formed by loss of a molecule of water. Such a linkage between two monosaccharide units through oxygen atom is called glycosidic linkage.
Section – E
Question 33.
(a) (I) Account for the following: 3+2
(i) E° value for Mn
3+
/Mn
2+
couple is much more positive than that for Cr
3+
/Cr
2+
.
Answer:
The large positive E° value for Mn
3+
/Mn
2+
shows that Mn
2+
is much more stable than Mn
3+
due to stable half filled configuration (3d
5
). Therefore the 3
rd
ionisation energy of Mn will be very high and Mn
3+
is unstable and can be easily reduced to Mn
2+
. E° value for Cr
3+
/Cr
2+
is positive but small, i.e., Cr
3+
can also be reduced to Cr
2+
but less easily. Thus, Cr
3+
is more stable than Mn
3+
.
(ii) Sc
3+
is colourless whereas Ti
3+
is coloured in an aqueous solution.
Answer:
Sc
3+
is colourless due to absence of unpaired electrons in its ion (3d
0
4s
0
)
while Ti
3+
(3d
1
4s
0
) is coloured due to d-d transition of unpaired electron in it.
(iii) Actinoids show wide range of oxidation states.
Answer:
Because of very small energy gap between 5f, 6d and 7s subshells all their electrons can take part in bonding and show variable oxidation states.
(II) Write the chemical equations for the preparation of KMnO
4
from MnO
2
.
Answer:
Potassium Permangante (KMnO
4
) is prepared from pyrolusite ore (MnO
2
). The ore (MnO
2
) is fused with an alkali metal hydroxide like KOH in the presence of air or an oxidising agent like KNO
3
to give dark green potassium manganate (K
2
MnO
4
). K
2
MnO
4
disproportionates in a neutral or acidic solution to give potassium permanganate.
Or
(b)
(I) Account for the following: [2 + 2 + 1]
(i) Transition metals form alloys.
Answer:
The transition metals are quite similar in size and therefore, the atoms of one metal can substitute the atoms of other metal in its crystal lattice. Thus, on cooling a mixture solution of two or more transition metals solid alloys are formed.
(ii) Ce
4+
is a strong oxidising agent.
Answer:
Ce
4+
is a strong oxidizing agent because it tends to come in its stable oxidation state of +3.
(II) Write one similarity and one difference between chemistry of Lanthanoids and Actinoids.
Answer:
Similarity. Both lanthanoids and actìnoids show contraction in size and irregularity in their electronic configuration.
Difference. Actinoids show wide range of oxidation states but lanthanoids do not.
(III) Complete the following ionic equation:
Cr
2
\(\mathrm{O}_7^{2-}\) + 2OH
–
→
Answer:
Cr
2
\(\mathrm{O}_7^{2-}\) + 2OH
–
→ 2Cr\(\mathrm{O}_4^{2-}\) + H
2
O
Question 34.
(a) (I) Give reasons: [3 + 2]
(i) Aniline on nitration gives good amount of m-nitroaniline, though —NH
2
grouP is o/p directing in electrophilic substitution reactions.
Answer:
Because of nitration in an acidic medium, aniline gets protonated to give anilinium ion which is m-directing.
(ii) (CH
3
)
2
NH is more basic than (CH
3
)
3
N in an aqueous solution.
Answer:
Due to more steric hindrance in (CH
3
)
3
N it is less basic than (CH
3
)
2
NH.
(iii) Ammonolysis of alkyl halides is not a good method to prepare pure primary amines.
Answer:
Because the primary amine formed by ammonolysis itself acts as a nucleophile and produces further 2° and 3° alkyl amine.
(II) Write the reaction involved in the following:
(i) Carbyl amine test
Answer:
Aliphatic and aromatic primary amines on heating with chloroform and ethanolic KOH form isocyanides or carbylamines which are foul smelling substances. This reaction is known as carbylamines reaction.
(ii) Gabriel phthalimide synthesis
Answer:
Gabriel phthalirnide synthesis. It is used to prepare 1° amine (Primary amine). The starting compound is a phthalimide. But aromatic primary amines cannot be prepared by this method because aryl halides do not undergo nucleophilic substitution with the anion formed by phthalimide.
Example:
Or
(b) (I) Write the structures of A, B and C in the following reactions:
(II) Why aniline does not undergo Friedel-Crafts reaction?
(III) Arrange the following in increasing order of their boiling point:
C
2
H
5
OH, C
2
H
5
NH
2
, (C
2
H
5
)
3
N
Answer:
As a result, N of aniline acquires positive charge and hence it acts as a strong deactivating group for electrophilic substitution reaction. Consequently, aniline does not undergo Friedel Craft reaction.
(III) Increasing order of boiling point: (C 2 H 5 ) 3 N < C 2 H 5 NH 2 < C 2 H 5 OH
Question 35.
(a) Conductivity of 2 × 10
-3
M methanoic acid is 8 × 10
-5
S cm
-1
. Calculate its molar conductivity and degree of dissociation if 4 for methanoic acid is 404 S cm
2
mol
-1
.
Answer:
Given, Concentration, C = 2 × 10
-3
M ; Conductivity, K = 8 × 10
-5
S cm
-1
\(\Lambda_{\mathrm{m}}^{\mathrm{o}}\) = 404 S cm
2
mol
-1
Using formula,
Λ
m
= \(\frac{K \times 1000}{C}\) = \(\frac{8 \times 10^{-5} \times 1000}{2 \times 10^{-3}}\) S cm
2
mol
-1
= 4 × 10
-5+3
× 1000 = 4 × 10
-2
× 10
3
∴ Molar Conductivity, Λ
m
= 40 S cm
2
mol
-1
Now, Degree of dissociation, α = \(\frac{\Lambda_{\mathrm{m}}}{\Lambda_{\mathrm{m}}^{\circ}}\) = \(\frac{40}{404}\) = 0.099
(b) Calculate the ∆
r
G° and log K
c
for the given reaction at 298 K: [3 + 2]
Answer:
SET II Code No. 56/2/2
Except for the following questions, all the remaining questions have been asked in Set I.
Question 1.
Which of the following belongs to the class of Vinyl halides? [1]
Answer:
(b)
In Vinyl halide, halogen atom is attached to the sp 2 carbon of C = C bond.
Question 2.
What is the secondary valency of Cobalt in [Co(en
2
)Cl
2
]
+
? [1]
(a) 6
(b) 4
(c) 2
(d) 8
Answer:
(a) 6
Secondary valencies have characteristic spatial arrangements in coordination polyhedra and represent number of donor atoms in ligands.
Question 3.
When Benzene diazonium chloride reacts with phenol, it forms a dye. This reaction is called [1]
(a) Diazotisation reaction
(b) Condensation reaction
(c) Coupling reaction
(d) Acetylation reaction
Answer:
(c) Coupling reaction
Benzene diazonium chloride reacts with phenol in a basic medium and gives p-Hydroxyazobenzene. This is an example of the coupling reaction.
Question 4.
The slope in the plot of IRI vs. time for a zero order reaction is [1]
(a) \(\frac{+k}{2.303}\)
(b) -k
(c) \(\frac{-k}{2.303}\)
(d) +k
Answer:
(b) -k
Question 5.
Proteins are polymers of [1]
(a) Nucleic acids
(b) Amino acids
(c) Monosaccharides
(d) Amines
Answer:
(b) Amino acids
Question 6.
Retention of configuration is observed in [1]
(a) S
N
1 reaction
(b) S
N
2 reaction
(c) Neither S
N
1 nor 5
N
2 reaction
(d) S
N
2 reaction as well as S
N
1 reaction
Answer:
(a) S
N
1 reaction
Question 7.
An azeotropic mixture of two liquids will have a boiling point lower than either of the two liquids when it [1]
(a) shows a negative deviation from Raoult’s law.
(b) forms an ideal solution.
(c) shows a positive deviation from Raoult’s law.
(d) is saturated.
Answer:
(c) shows a positive deviation from Raoult’s law.
Question 9.
Which of the following does not give Cannizaro reaction?
(a) (CH
3
)
3
C—CHO
(b) (CH
3
)
2
CH—CHO
(c)
(d) HCHO
Answer:
(b) (CH
3
)
2
CH—CHO
Aldehydes which have α-Hydrogen atom do not undergo Cannizaro reaction.
Question 11.
Aldehydes and ketones react with hydroxylamine to form [1]
(a) hydrazones
(b) cyanohydrins
(c) semicarbazopes
(d) Oxime
Answer:
(d) Oxime
Question 12.
For a reaction 2A → 3B, rate of reaction \(-\frac{d[\mathbf{A}]}{d t}\) is equal to [1]
(a) \(\frac{+3}{2} \frac{d[\mathrm{~B}]}{d t}\)
(b) \(\frac{+2}{3} \frac{d[\mathrm{~B}]}{d t}\)
(c) \(\frac{+1}{3} \frac{d[\mathrm{~B}]}{d t}\)
(d) \(+\frac{2 d[\mathrm{~B}]}{d t}\)
Answer:
(b) \(\frac{+2}{3} \frac{d[\mathrm{~B}]}{d t}\)
For 2A → 3B
Rate of reaction \(\frac{-1}{2} \frac{d[\mathrm{~A}]}{d t}\) = \(\frac{1}{3} \frac{d[\mathrm{~B}]}{d t}\)
⇒ \(\frac{-d[\mathrm{~A}]}{d t}\) = \(\frac{2}{3} \frac{d[\mathrm{~B}]}{d t}\)
Question 14.
Which one among the following metals of 3d series has the lowest melting point? [1]
(a) Fe
(b) Mn
(c) Zn
(d) Cu
Answer:
(c) Zn
Zinc has the minimum melting point among 3d series of transition metals, because of absence of unpaired d-electrone.
Given below are two statements labelled as Assertion (A) and Reason (R). Select the most appropriate answer from the options given below:
(a) Both (A) and (R) are true and (R) is the correct explanation of (A).
(b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
(c) (A) is true, but (R) is false.
(d) (A) is false, but (R) is true.
Question 15.
Assertion (A) : Elevation in boiling point is a colligative property. [1]
Reason (R) : The lowering of vapour pressure of solution causes elevation in boiling point.
Answer:
(b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
Elevation in boiling point is a colligative property as it is dependent on the solute- solvent ratio or number of particles. When a non-volatilq solute is added in a solution then its vapour pressure decreases than that of pure solvent and more heat has to be supplied to the solution to boil it.
Question 16.
Assertion (A) : Chlorobenzene is resistant to electrophilic substitution reaction. [1]
Reaction (R) : C — Cl bond in chlorobenzene acquires partial double bond characters due
to resonance.
Answer:
(d) (A) is false, but (R) is true.
Chlorobenzene is resistant to nucleophilic substitution reaction.
Question 18.
Assertion (A) : Transition metals have high enthalpy of atomisation. [1]
Reason (R) : Greater number of unpaired electrons in transition metals results in weak metallic bonding. [1]
Answer:
(c) (A) is true, but (R) is false.
High effective nuclear charge and a large number of valence electrons in transition metals cause very strong metallic bonding which results in high enthalpies of atomisation.
Question 23.
(a) What is the difference between a nucleoside and nucleotide?
Answer:
Nucleoside. A nucleoside contains only two basic components of nucleic acids, i.e., a pentose sugar and a nitrogenous base. It is formed by the attachment of a base to 1′ position of sugar.
Nucleotides. A nucleotide contains all the three basic components of nucleic acids, i.e., a phosphoric acid group, a pentose sugar and a nitrogenous base. These are formed by the esterification of C 5 —OH of the sugar of the nucleoside with phosphoric acid.
(b) What products would be formed when a nucleotide from DNA containing thymine is hydrolysed? [2 × 1]
Answer:
When nucleotide from DNA containing thymine β-D-2-deoxyribose and phosphoric acid are obtained as products.
Question 24.
Write the chemical equation involved in the following: [2 × 1]
(a) Kolbe’s reaction
Answer:
Kolbe’s reaction. Phenol reacts with CO
2
in presence of sodium hydroxide (NaOH) at 4 – 7 Atm and 390 — 410 K giving salicvlic acid.
Example:
(b) Williamson synthesis
Answer:
Williamson synthesis. The reaction involves the nucleophilic substitution of the halide ion from the alkyl halide by the alkoxide ion by S
N
2 mechanism.
Example:
SET III Code No. 56/2/3
Except for the following questions, all the remaining questions have been asked in Set I and Set II.
Question 1.
Auto oxidation of chloroform in air and sunlight produces a poisonous gas known as
(a) Tear gas
(b) Mustard gas
(c) Phosgene gas
(d) Chlorine gas
Answer:
(c) Phosgene gas
Question 2.
Which of the following ligands is an ambidentate ligand? [1]
(a) CO
(b) NO
2
(c) NH
3
(d) H
2
O
Answer:
(b) NO
2
NO 2 can donate electron from either oxygen or nitrogen atom of its molecule and forms O — N — O and — NO 2 linkage.
Question 3.
Among the following, which has the highest value of pK
b
? [1]
Answer:
(d)
Electron withdrawing group like —NO 2 destabilises the anilinium cation and decreases basic strength and increases pK b value.
Question 4.
The slope in the plot of log \(\frac{[R]_0}{[R]}\) vs. time for a first order reaction is [1]
(a) \(\frac{+k}{2.303}\)
(b) +k
(c) \(\frac{-k}{2.303}\)
(d) -k
Answer:
(a) \(\frac{+k}{2.303}\)
Question 5.
When D-glucose reacts with HI, it forms
(a) Gluconic acid
(b) n-hexane
(c) Saccharic acid
(d) Iodohexane
Answer:
(b) n-hexane
Question 6.
Inversion of configuration occurs in
(a) S
N
2 reaction
(b) S
N
1 reaction
(c) Neither S
N
2 nor S
N
1 reaction
(d) S
N
1 as well as S
N
2 reaction
Answer:
(a) S
N
2 reaction
S N 2 mechanism requires the attack of Nu – from the backside of ‘C’ atom. So the product assumes a stereochemical position opposite to the leaving group originally occupied and this is K/a inversion of configuration.
Question 7.
Solubility of gas in liquid decreases with increase in
(a) Pressure
(b) Temperature
(c) Volume
(d) Number of solute molecules
Answer:
(b) Temperature
Increased temperature causes an increase in K.E. which breaks intermolecular bonds and gas escapes the solution.
Question 8.
Which of the following relations is incorrect?
(a) R = \(\frac{1}{k}\left(\frac{l}{a}\right)\)
(b) G = \(k\left(\frac{a}{l}\right)\)
(c) G = \(k\left(\frac{l}{a}\right)\)
(d) \(\wedge_m\) = \(\frac{k}{c}\)
Answer:
(c) G = \(k\left(\frac{l}{a}\right)\)
Question 9.
The reagent that can be used to distinguish acetophenone and benzophenone is
(a) 2,4-dinitrophenyl hydrazine
(b) aqueous NaHSO
3
(c) Fehling solution
(d) I
2
and NaOH
Answer:
(d) I
2
and NaOH
Benzophenone has benzene rings while acetophenone has linear structure. Thus, acetophenone easily gives iodoform test to produce yellow ppt. while benzophenone cannot do so.
Question 11.
Which of the following compounds will undergo self-condensation in the presence of dilute NaOH solution? [1]
(a) C
6
H
5
CHO
(b) CH
3
CH
2
CHO
(c) (CH
3
)
3
C—CHO
(d) H—CHO
Answer:
(b) CH
3
CH
2
CHO
Due to presence of α-Hydrogen atom CH 3 CH 2 CHO undergoes self condensation.
Question 13.
Which of the following transition metals shows +1 and +2 oxidation states? [1]
(a) Mn
(b) Zn
(c) Sc
(d) Cu
Answer:
(d) Cu
Question 14.
The formula of the complex Iron (III) hexacyanidoferrate (II) is: [1]
(a) Fe
2
[Fe(CN)
6
]
3
(b) Fe
4
[Fe(CN)
6
]
3
(c) Fe[Fe(CN)
6
]
(d) Fe
3
[Fe(CN
6
]
2
Answer:
(b) Fe
4
[Fe(CN)
6
]
3
Given below are two statements labelled as Assertion (A) and Reason (R). Select the most appropriate answer from the options given below:
(a) Both (A) and (R) are true and (R) is the correct explanation of (A).
(b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
(c) (A) is true, but (R) is false.
(d) (A) is false, but (R) is true.
Question 15.
Assertion (A): The enthalpy of mixing ∆
mix
H is equal to zero for an ideal solution. [1]
Reason (R) : For an ideal solution the interaction between solute and solvent molecules is stronger than the interactions between solute-solute or solvent-solvent molecules.
Answer:
(c) (A) is true, but (R) is false.
For ideal solution the interaction between solute and solvent molecules is weaker than solute-solute or solvent-solvent interaction.
Question 16.
Assertion (A): Molar conductivity decreases with increase in concentration. [1]
Reason (R) : When concentration approaches zero, the molar conductivity is known as limiting molar conductivity.
Answer:
(d) (A) is false, but (R) is true.
Molar conductivity decreases with decrease in concentration.
Question 17.
Assertion (A): Transition metals show their highest oxidation state with oxygen. [1]
Reason (R) : The ability of oxygen to form multiple bonds to metals.
Answer:
(a) Both (A) and (R) are true and (R) is the correct explanation of (A).
Question 18.
Assertion (A): Chlorobenzene is resistant to nucleophilic substitution reaction at room temperature.
Reason (R) : C — Cl bond gets weaker due to resonance. [1]
Answer:
(c) (A) is true, but (R) is false.
The C — Cl bond gets stronger due to resonance and the double character acquired by it due to delocalization of electrons in the ring.
Question 19.
What are nucleic acids? Why two strands in DNA are not identical but are complementary? [1 × 2]
Answer:
Nucleic acids are biologically important polymers of proteins containing chromosomes which are present in all living cells. The two strands of DNA are said to be complementary to each other in the sense that the sequences of bases in one strand automatically determine that of the other. For example, whenever adenine (A) appears in one strand, a thymine (T) appears opposite to it in the other strand.
Question 20.
Do the following conversions in not more than two steps: [2 × 1]
(a) CH
3
COOH to CH
3
COCH
3
Answer:
Ethanoic acid to propanone
(b)
Answer:
